BASIS OF SIMULTANEOUS HECKE EIGENFORMS 1. INTRODUCTION

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1 BASIS OF SIMULTANEOUS HECKE EIGENFORMS MARIA HEMPEL 1. INTRODUCTION The aim of this presentation is to show that there exist bases of simultaneous Hecke eigenforms (i.e., bases consisting of functions, which are eigenforms to all Hecke operators T n ) both of the space of cusp forms S k (Γ) and of the space of modular forms M k (Γ). To that end we will define a scalar product on S k (Γ), called the Petersson Inner Product, and show that all Hecke operators are hermitian with respect to that product. The rest will follow readily via some linear algebra. We assume familiarity with the definition of Hecke operators and Poincaré series, with their basic properties and with the corresponding notaions such as j γ (z) (cz + d) k (these have been introduced in previous talks).. CONSTRUCTION OF THE PETERSSON INNER PRODUCT Usually we define an inner product on spaces of functions via the use of the integral < f, g > f (z)g(z)c(z)dξ(z) for some region on which f and g are defined, some function c(z) and some measure dξ(z). Since we are trying to define the inner product on S k (Γ) we are dealing with functions wich are wholly defined by their values on any fundamental domain. We therefore set :. Now we have to choose a Γ -invariant integrand and measure. Let s first try to define the function c(z) such that the integrand becomes invariant under Γ. If we compute, f (γz)g(γz) j γ (z) f (z)j γ (z)g(z) j γ (z) f (z)g(z) using Im(γz) j γ (z) k Im(z),we see that, f (γz)g(γz)im(γz) k f (z)g(z)im(z) k 1

2 MARIA HEMPEL and thus set c(z) c(x + iy) : y k. To find the measure dξ(z), we guess that it is dξ(z) dξ(x + iy) y (It is in fact the Haar measure). We now show that it is invariant under the action of Γ : dξ(γz) Im(γz) d(γz) dz Im(γz) detγ j γ (z) k Im(γz) 1 j γ (z) 4 k Im(z) Finally, we have to check that our potential inner product is indeed well defined. As we saw in a previous talk f (z)y k is bounded for any function f in the space of cusp forms. By consequence f, g S k (Γ) C R + such that C f (z)g(z)y k dξ(z) f (z)g(z)y k dξ(z) 1 dξ(z) C 1 y dydx C π 1 x 3 The Integral is thus bounded for any two functions in the space of cusp forms and we are now free to define the Definition (Petersson Inner Product). For any two functions f and g in S k (Γ) we define the Petersson Inner Product as follows: < f, g > f (z)g(z)y k y Notice that the Petersson Inner Product is an indeed an inner product. Recognize also that S k (Γ) is a finite dimensional normed vector space and therefore a Hilbert space. 3. COMPLETENESS OF THE POINCARÉ SERIES In this paragraph we want to see, that the Poincaré series are generators of S k (Γ). To that end we state the following lemma: Lemma. Let f S k (Γ) and P k m be the Poincaré series of weight k with m 1. Then < f, P k m > C k,m a m

3 BASIS OF SIMULTANEOUS HECKE EIGENFORMS 3 where C k,m and a m is the m-th Fourier coefficient of f. Γ(k 1) (4πm) k 1 The proof is a calculation involving the Γ - invariance of the integral defining the Petersson Inner Product, identities of the automorphy factor studied in previous talks and unfolding of the integral. For a detailed proof see Gunning s Lectures on Modular Forms. We can now prove the following Theorem. For k > the set G : {P k m(z) m 1} generates S k (Γ). which is what we wanted. Proof. Let M be the subspace of S k (Γ) generated by G, m 1 and k >. Then f M : < f, P k m > 0 and using the above lemma we see that C k,m a m 0 Thus we see that a m 0 amounting to f 0, meaning that M {0} 4. HECKE OPERATORS ARE HERMITIAN In this section we show the following Theorem. Hecke Operators are hermitian with respect to the Petersson Inner Product. To do so we start with Step 1: Some identities For T n P m (z) n k P m (z) l1 l1 assume the following symmetry conditions: (1) () (3) with C R These may be proven by using that c l (n,m)e(lz) c m (l)e(lz) c l (m,n) c l (n,m) m 1 k c m (l,n) n 1 k c n (l,m) ( ) nl < T n P l, P m > C d 1 k c m d d (n,l) b(l) d (n,l) d k 1 a ( ) ln d

4 4 MARIA HEMPEL for f l1 a(l)e(lz) and T n f l1 b(l)e(lz), wich was proven by the previous group. Sart by showing (3) using the above lemma and then proceed to prove (1) and () by using (3). Step : Lemma. (4) < T n P m, P q >< T n P q, P m > Proof. Recall from above that < f, P m > C k,m a m Γ(k 1) (4π) k 1 m1 k a m : Am 1 k a m Using (1) and () we can therefore compute the following: < T n P m, P q > Aq 1 k a q An k q 1 k c q (m,n) An k n 1 k c n (m,q) An k n 1 k c n (q,m) An k m 1 k c m (q.n) < T n P q, P m > Step 3: We can finally prove the theorem Proof. Let f N n1 a np n andg M m1 a mp m be two cusp forms. First observe that < P n, T k P m > R. This follows from (3) and < P a, P b >< T 1 P a, P b >< T 1 P b, P a >< P b, P a > With this and (4) we can therefore compute the following: < T l f, g >< T l N n1 a n P n, N M n1 m1 N M n1 m1 M N m1 n1 N M m1 a m P m > a n b m < T l P n, P m > a n b m < T l P m, P n > a n b m < P n, T l P m > M < a n P n, T l b m P m > n1 m1 < f, T l g >

5 BASIS OF SIMULTANEOUS HECKE EIGENFORMS 5 5. HARVEST Theorem. For every positive integer n there exists a basis of eigenforms of T n for S k (Γ). This hodls since by Linear Algebra there is an orthonormal basis of eigenvectors for every hermitian liner operator on a finite dimensional vector space. Theorem. There is a basis of simulaneous Hecke eigenforms for S k (Γ). This also follows from a theorem of Linear Algebra, requiring the Hermitian operator to commute. This was indeed proven by the previous group. Corollary. There is a basis of simultaneous eigenforms for M k (Γ)